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Question

If $$\displaystyle x\sin ^{3}0 +y\cos ^{3}0 =\sin 0 \cos 0 \: \: and\: \: x\sin 0 =y\cos 0 $$ then


A
x3+y3=1
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B
x2y2=1
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C
x2+y2=1
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D
x3y3=1
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Solution

The correct option is D $$\displaystyle x^{2}+y^{2}=1$$
Given: $$\displaystyle x\sin ^{3}0 +y\cos ^{3}0 =\sin 0 \cos 0 \: \: and\: \: x\sin 0 =y\cos 0 $$ 
Now, $$\displaystyle x\sin ^{3}0 +y\cos ^{3}0 =\sin 0 \cos 0$$
$$\displaystyle x\sin 0. \sin^2{3}0 +y\cos ^{3}0 =\sin 0 \cos 0$$
$$\displaystyle y \cos 0. \sin^2{3}0 + \cos ^{3}0 =\sin 0 \cos 0$$ (Given, $$x\sin 0 =y\cos 0 $$ )
$$\displaystyle y \cos 0(\sin^2{3}0 + \cos ^{2}0) =\sin 0 \cos 0$$ (Since, $$sin^2\theta + cos^2 \theta = 1$$)
$$\displaystyle y  =\sin 0$$
Similarly, $$\displaystyle x  =\cos 0$$
Now, $$x^2 + y^2 = sin^20+ cos^2 0 = 1$$ (Since, $$sin^2\theta + cos^2 \theta = 1$$) 

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